Question

One mole of a vander Waal's gas obeying the equation
$(p+\frac{a}{V^2})(V-b)=RT$
undergoes the quasi-static cyclic process which is shown in the p-V diagram. The net heat absorbed by the gas in this process is:

• A. $\frac{1}{2}(p_1-p_2)(V_1-V_2)$
• B. $\frac{1}{2}(p_1+p_2)(V_1-V_2)$
• C. $\frac{1}{2}(p_1+\frac{a}{V_1^2}-p_2-\frac{a}{V_2^2})(V_1-V_2)$
• D. $\frac{1}{2}(p_1+\frac{a}{V_1^2}+p_2+\frac{a}{V_2^2})(V_1-V_2)$

Answer 1

The correct option is A $\frac{1}{2}(p_1-p_2)(V_1-V_2)$

Workdone during the complete cycle is equal to the area enclosed by the P-V graph

$\therefore$ $W=\frac{1}{2}\times \left(OB\right)\times \left(OA\right)=\frac{1}{2}(p_1-p_2)(V_1-V_2)$

Also the change in internal energy in a complete cycle is Zero i.e $\Delta U=0$

Using first law of thermodynamics : $Q=\Delta U+W$

$\therefore$ $Q=0+W=W$

Thus the heat absorbed by the gas in a complete cycle $Q=W=\frac{1}{2}(p_1-p_2)(V_1-V_2)$